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a. Suppose the lifetime \(X\) of a component, when measured in hours, has a gamma distribution with parameters \(\alpha\) and \(\beta\). Let \(Y=\) the lifetime measured in minutes. Derive the pdf of \(Y\). [Hint: \(Y \leq y\) iff \(X \leq y / 60\). Use this to obtain the cdf of \(Y\) and then differentiate to obtain the pdf.] b. If \(X\) has a gamma distribution with parameters \(\alpha\) and \(\beta\), what is the probability distribution of \(Y=c X\) ?

Short Answer

Expert verified
a. The pdf of Y is \(f_Y(y) = \frac{1}{60} f_X\left(\frac{y}{60}\right)\). b. Y has a gamma distribution with parameters \(\alpha\) and \(\frac{\beta}{c}\).

Step by step solution

01

Understanding the Transformation

The exercise involves transforming the random variable from hours to minutes. If the lifetime in hours is denoted by the random variable \(X\), then the lifetime in minutes \(Y\) would be \(Y = 60X\). We will use this relationship to transform the distribution of \(X\) to find the distribution of \(Y\).
02

Cumulative Distribution Function of Y

To find the cumulative distribution function (CDF) of \(Y\), note that \(Y \leq y\) if and only if \(X \leq \frac{y}{60}\). The CDF of \(X\) is denoted by \(F_X(x)\). Thus, the CDF of \(Y\) is given by \(F_Y(y) = P(Y \leq y) = P\left(X \leq \frac{y}{60}\right) = F_X\left(\frac{y}{60}\right)\).
03

Differentiate to find the PDF of Y

The probability density function (PDF) of \(Y\), denoted by \(f_Y(y)\), is the derivative of the CDF. Therefore, \(f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} F_X\left(\frac{y}{60}\right)\). Using the chain rule, \(f_Y(y) = f_X\left(\frac{y}{60}\right) \cdot \frac{1}{60}\), where \(f_X(x)\) is the PDF of the gamma distribution.
04

Probability Distribution of cX

Let \(Z = cX\). If \(X\) has a gamma distribution with parameters \(\alpha\) and \(\beta\), transforming \(X\) by a constant \(c\) results in \(Z\) also following a gamma distribution. The parameters are affected as follows: \(Z\) has a gamma distribution with parameters \(\alpha\) and \(\frac{\beta}{c}\). Hence, the scale parameter \(\beta\) is divided by the constant \(c\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function (PDF)
The Probability Density Function (PDF) is a fundamental concept in probability and statistics. It describes how the probability of a continuous random variable is distributed over its possible values. In simpler terms, the PDF tells you the likelihood of a random variable taking on a particular value in a continuous range. For the gamma distribution, the PDF is defined using the shape parameter \(\alpha\) and the scale parameter \(\beta\). The formula looks like this: \[ f_X(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \]where \(\Gamma(\alpha)\) is the gamma function. In the exercise, we transformed a random variable from hours to minutes, leading to the formation of a new PDF, \(f_Y(y)\), by relation: \[ f_Y(y) = f_X\left(\frac{y}{60}\right) \cdot \frac{1}{60} \].Here, we used the chain rule of differentiation to obtain the PDF of \(Y\). This step is pivotal because it transforms the PDF from its original unit to the new unit while preserving the probability structure.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) represents the probability that a random variable takes on a value less than or equal to a certain number. For a gamma-distributed variable \(X\), this would be expressed as \(F_X(x) = P(X \leq x)\). The CDF is beneficial because it gives a sense of the cumulative probability accumulation up to a point. To find the CDF of the transformed variable \(Y = 60X\), we used the transformation hint: \(Y \leq y\) if and only if \(X \leq y/60\). Therefore, the CDF of \(Y\) is \(F_Y(y) = F_X(y/60)\), allowing us to connect the random variable in hours to minutes. This CDF provides the basis for further finding the PDF through differentiation, linking continuous probability across transformations.
Transformation of Random Variables
Transforming random variables is an essential operation when dealing with different measurement scales or units. In this exercise, the lifetime of a component measured in hours was transformed to minutes. This is mathematically represented by \(Y = 60X\), where \(X\) represents the original random variable and \(60\) is the transformation constant. Such transformations allow us to express probabilities in the new units of measure while adhering to original statistical properties. When transforming distributions like the gamma distribution, not only do we need to adjust the values, but also consider how kernels such as the PDF and CDF behave. By differentiating the transformed CDF, we elegantly derived the new PDF for \(Y\). This shows us how versatile probability distributions are through proper transformations.
Scale Parameter
The scale parameter \(\beta\) is a critical element of the gamma distribution, which affects the spread or scaling of the distribution's PDF and CDF. The scale parameter essentially stretches or compresses the distribution along the x-axis. When we transform a random variable like \(X\) with a gamma distribution and change it by a constant \(c\), creating \(Z = cX\), the new gamma distribution of \(Z\) sees its scale parameter adjusted to \(\beta/c\). This means the scale parameter is inversely proportional to the constant \(c\)—dividing \(\beta\) by \(c\) stretches the distribution wider, indicating a longer spread of values. Understanding the scale parameter is crucial for making precise probability assessments in different transformational scenarios, keeping the probabilistic behavior intact although in a different measure.

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Most popular questions from this chapter

Let \(X\) denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article "Blade Fatigue Life Assessment with Application to VAWTS" (J. of Solar Energy Engr., 1982: 107-111) proposes the Rayleigh distribution, with pdf $$ f(x ; \theta)=\left\\{\begin{array}{cc} \frac{x}{\theta^{2}} \cdot e^{-x^{2} \cdot\left(2 \theta^{2}\right)} & x>0 \\ 0 & \text { otherwise } \end{array}\right. $$ as a model for the \(X\) distribution. a. Verify that \(f(x ; \theta)\) is a legitimate pdf. b. Suppose \(\theta=100\) (a value suggested by a graph in the article). What is the probability that \(X\) is at most 200? Less than 200? At least 200? c. What is the probability that \(X\) is between 100 and 200 (again assuming \(\theta=100\) )? d. Give an expression for \(P(X \leq x)\).

The error involved in making a certain measurement is a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cc} .09375\left(4-x^{2}\right) & -2 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Compute \(P(X>0)\). c. Compute \(P(-1.5)\).

The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean value \(40 \mathrm{~V}\) and standard deviation \(1.5 \mathrm{~V}\). a. What is the probability that the voltage of a single diode is between 39 and 42 ? b. What value is such that only \(15 \%\) of all diodes have voltages exceeding that value? c. If four diodes are independently selected, what is the probability that at least one has a voltage exceeding 42 ?

Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential36. Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential is well known. The is well known. The paper "Effects of 2,4-D Formulation and Quinclorac on Spray Droplet Size and Deposition" (Weed Technology, 2005: 1030-1036) investigated the effects of herbicide formulation on spray atomization. A figure in the paper suggested the normal distribution with mean \(1050 \mu \mathrm{m}\) and standard deviation \(150 \mu \mathrm{m}\) was a reasonable model for droplet size for water (the "control treatment") sprayed through a \(760 \mathrm{ml} / \mathrm{min}\) nozzle. a. What is the probability that the size of a single droplet is less than \(1500 \mu \mathrm{m}\) ? At least \(1000 \mu \mathrm{m}\) ? b. What is the probability that the size of a single droplet is between 1000 and \(1500 \mu \mathrm{m}\) ? c. How would you characterize the smallest \(2 \%\) of all droplets? d. If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds \(1500 \mu \mathrm{m}\) ?

Consider the pdf for total waiting time \(Y\) for two buses $$ f(y)=\left\\{\begin{array}{cl} \frac{1}{25} y & 0 \leq y<5 \\ \frac{2}{5}-\frac{1}{25} y & 5 \leq y \leq 10 \\ 0 & \text { otherwise } \end{array}\right. $$ introduced in Exercise 8. a. Compute and sketch the cdf of \(Y\). [Hint: Consider separately \(0 \leq y<5\) and \(5 \leq y \leq 10\) in computing \(F(y)\). A graph of the pdf should be helpful.] b. Obtain an expression for the \((100 p)\) th percentile. [Hint: Consider separately \(0

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